1Pritish that is not the answer man
answer is e100
can anyone explain?????????how
find the maximum product(approximate) of positive real numbers whose sum is 271 and also prove the result.
try it very good question and very pretty answer...
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39Let the real numbers be x and y.
We are given that
x + y = 271.
We have to maximise xy
Let xy = p
=> x(271 - x) = p
Differentiate wrt x.
271 - 2x = dp/dx
Put dp/dx = 0.
271 = 2x
=> x = 271/2
So y = 271/2.
Test by the double derivative.
d²p/dx² = -2 < 0
This implies x = 271/2 is a point of maxima.
So our real numbers are 271/2, 271/2.
This can also be proved by an axiom in algebra which states that -:
"When the sum of positive numbers is given to be constant, the product of these numbers is maximum when all the numbers are equal"
So your product is (271/2)² = 18360.25
1
62Lets assume that this is made of n numbers
then the maximum value will be when all of them are equal..
This part would be eas;y to prove using AM GM inequality
Hence the number would be (271/n)^n
So the aim now is to maximize (271/n)^n when n is an integer..
Anyone who wants to fill in the gaps from the proof?
1it is like this
USing inequaliy theorem for maxima and minima
i.e ,
IF a1 + a2 + ............... + an = k (constant )
for the value of a1.a2........an to be maximum a1 = a2 = ......... = an and maximum value of a1.a2........an is \left(\frac{k}{n} \right)^{n}
so we got here
a1 + a2 = 271
hence for maximum product we have a1.a2 = \left(\frac{271}{2} \right)^{2}
hence the answer ...........
1may be the answer is wrong or may be e100 =( 271/2)2 which i don't thnk is
and my answer is as stated by pritish [1]
62manmay, you have taken n =2
Why?
I mean after a step you have missed out the fact that n is not equal to 2
1bhaiya frm the inequality theorem na
as i stated which was given in the book
62\\\frac{d}{dn}\left( \frac{271}{n}\right)^n=\left( \frac{271}{n}\right)^nln(271/en)
Which is zero when 271=en
So 271/e =n
But Here what we have to keep in mind is that n is an integer..
so n = 100 will be the most appropriate integer closest to 271/e
So the maxima will be 2.71100
1i still don't get it
using inequality theorem as i stated abv in post #6
maximum value is \left(\frac{k}{n} \right)^{n} itself so why we further maximise it NISHANT BHAIYA ??
62yup gallardo i realised it at home after leaving office :P
Yesterday was "ALL India Bengal Band" called by CPI .. so couldnt come to office :D
at manmay.. no that is the value.. we have to maximize it.
It has not been menhtioned anywhere in the question that n=2
We have to maximize this given value over all values of n!
(What you are doing is that this is the maximum value for a fixed value of n...)
After that even n can vary
341Two links from the past (in another Avatar!)
http://www.goiit.com/posts/list/algebra-write-271-as-the-sum-of-positive-real-numbers-so-as-73541.htm#364087
and
http://www.goiit.com/posts/list/algebra-a-problem-in-maximization-everyone-s-invited-94553.htm
1thnk u NISHANT BHAIYA............. I FINALLY UNDERSTOOD AND GOT MY MISTAKE [1]